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Rules of thumb for estimating visual anglesThat's a creative suggestion. Not sure I can execute it (and then move my arms to the right position) even remotely accurately. One can rule out the 46 degree halo with only one halving, or see if the diameter is 92 without any halving at all.

Calculating the resistance of a 3D shape between two pointsMy model above has about 10,000 nodes. It took about 100,000 iterations to converge. I hope I've done it right, some reverse engineering implies what I've been doing is solving Laplace's equation $\nabla^2 \phi = 0$ with boundary conditions $(\hat{n} \cdot \nabla) \phi = 0$ for non-contacts and $\phi = const$ for the contacts. This seems very plausible to me now. You've been very helpful, thanks.

Calculating the resistance of a 3D shape between two pointsI'd decided this approach was incorrect, though perhaps you could persuade me otherwise. The reason I had decided this was this: Lets say we use a rectangular grid. Now consider a relatively thin straight object. If the object, of length $l$ is aligned with the grid, it would contain $n$ resistors and have a resistance $R$, if it lies diagonally to the grid, then it would contain $~\sqrt{2}n$ resitors and have a resistance of $\sqrt{2}R$. Perhaps you are suggesting modifying the resistances as part of the relaxation, though I'm not sure how one would go about that.